(v) 1−i1+i\dfrac{1-i}{1+i}1+i1−i 1−i1+i=1−i1+i⋅1−i1−i=(1−i)21−i2=1−2i+i22=−2i2=−i\begin{aligned} \frac{1-i}{1+i} &=\frac{1-i}{1+i}\cdot\frac{1-i}{1-i} =\frac{(1-i)^2}{1-i^2}\\ &=\frac{1-2i+i^2}{2}=\frac{-2i}{2}=-i \end{aligned}1+i1−i=1+i1−i⋅1−i1−i=1−i2(1−i)2=21−2i+i2=2−2i=−i So z=−iz=-iz=−i. r=∣z∣=1,θ=−π2 r=|z|=1,\quad \theta=-\frac{\pi}{2}r=∣z∣=1,θ=−2π −i=cos(−π2)+isin(−π2)\boxed{-i=\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)}−i=cos(−2π)+isin(−2π)